Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6800 Accepted Submission(s): 2760
Problem Description
Consider equations having the following form:
a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.
Determine how many solutions satisfy the given equation.
a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.
Determine how many solutions satisfy the given equation.
Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
End of file.
End of file.
Output
For each test case, output a single line containing the number of the solutions.
Sample Input
1 2 3 -4
1 1 1 1
Sample Output
39088
0
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int f1[1000001]={0};
int f2[1000001]={0};
int main()
{
int a=0,b=0,c=0,d=0;
while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
{
memset(f1,0,sizeof(f1));
memset(f2,0,sizeof(f2));
if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0)
{
printf("0\n");
continue;
}
for(int i=1;i<=100;i++)
{
for(int j=1;j<=100;j++)
{
int k=a*i*i+b*j*j;
if(k>=0)
{
f1[k]++;
}
else
{
f2[-k]++;
}
}
}
int sum=0;
for(int i=1;i<=100;i++)
{
for(int j=1;j<=100;j++)
{
int k=c*i*i+d*j*j;
if(k>0)
{
sum+=f2[k];
}
else
{
sum+=f1[-k];
}
}
}
printf("%d\n",16*sum);
}
return 0;
}